The equation of hyperbola $H$ is $\dfrac {(x-4)^{2}}{16}-\dfrac {(y-4)^{2}}{64} = 1$. What are the asymptotes?
Answer: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y-4)^{2}}{64} = - 1 + \dfrac {(x-4)^{2}}{16}$ Multiply both sides of the equation by $64$ $(y-4)^{2} = { - 64 + \dfrac{ (x-4)^{2} \cdot 64 }{16}}$ Take the square root of both sides. $\sqrt{(y-4)^{2}} = \pm \sqrt { - 64 + \dfrac{ (x-4)^{2} \cdot 64 }{16}}$ $ y - 4 = \pm \sqrt { - 64 + \dfrac{ (x-4)^{2} \cdot 64 }{16}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y - 4 \approx \pm \sqrt {\dfrac{ (x-4)^{2} \cdot 64 }{16}}$ $y - 4 \approx \pm \left(\dfrac{8 \cdot (x - 4)}{4}\right)$ Add $4$ to both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{2}{1}(x - 4)+ 4$